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I know that an abelian triangulated category is semisimple,i.e.,any exact sequence splits.But Why does any distinguished triangle is isomorphic to a triangle of the form $X \stackrel{f}{\rightarrow} Y \stackrel{g}{\rightarrow} \operatorname{ker} f[1] \oplus \operatorname{coker} f \stackrel{h}{\rightarrow} X[1]$?(cf. Gelfand and Manin, Exercises to IV 1).

My try:Let $X \stackrel{f}{\rightarrow} Y \stackrel{g}{\rightarrow} Z \stackrel{h}{\rightarrow} X[1]$ be a distinguished triangle.Then we have $ \operatorname{coker} f \stackrel{g}{\rightarrow} Z \stackrel{h}{\rightarrow} \operatorname{ker} f[1]$.If it is exact at $\operatorname{coker} f,Z$,and $\operatorname{ker} f[1]$,then we have $Z\cong \operatorname{ker} f[1] \oplus \operatorname{coker} f$.But I can't show the exactness.

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  • $\begingroup$ If the category is semisimple, then split the map $f:X\to Y$ so that you have $X=Ker(f)\oplus X'$ and $Y=Coker (f)\oplus Y'$. Noe you can describe your map $f$ as being 0 on its kernel and an iso from $X'$ into Coker(f). The result now folllows from the fact that direct suma of tengles are rriangle $\endgroup$ – Marco Farinati Jun 12 at 23:06
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Split $X=Ker(f)\oplus X'$, and $Y\cong X'\oplus Coker(f)$. Now you can describe $f$ as a direct sum of 3 maps, zero in $Ker(f)$, identity on $X'$ and $0\to Coker(f)$. The result follows from the fact that identity fits into a triangle with 0 as cone, and the zero maps similarly (just traslate the triangle with identity)

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